Question

нужно решить 6,7,8,9,10 с пояснениями
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Answer 1

$6)\ \ a=64\ ,\ b=81\\\\\sqrt{\dfrac{a}{b}}-\sqrt{ab}=\sqrt{\dfrac{64}{81}}-\sqrt{64\cdot 81}=\dfrac{\sqrt{64} }{\sqrt{81} }-\sqrt{54} \cdot \sqrt{81}=\dfrac{8}{9}-8\cdot 9=\dfrac{8}{9}-72=-\dfrac{640}{9}$

$7)\ \ x=\sqrt{0,3}\ \,\ \ y=\sqrt{2,7}\\\\0,3xy-x^2=0,3\cdot \sqrt{0,3}\cdot \sqrt{2,7}- (\sqrt{0,3})^2=0,3\cdot \sqrt{0,3}\cdot \sqrt{9\cdot 0,3} -0,3=\\\\=0,3\cdot 0,3\cdot 3-0,3=3.6-0,3=3,3$

$8)\ \ m$

$9)\ \ \sqrt{(3,13^2-3,12^2)(8,2^2-1,8^2)}=\\\\=\sqrt{(3,13-3,12)(3,13+3,12)(8,2-1,8)(8,2+1,8)}=\\\\=\sqrt{0,01\cdot 6,25\cdot 6,4\cdot 10}=0,1\cdot 2,5\cdot \sqrt{64}=0,25\cdot 8=2$

$10)\ \ -3\leq b\leq -1\\\\\sqrt{b^2+10b+25}-\sqrt{b^2-12b+36}-\sqrt{25b^2}=\sqrt{(b+5)^2}-\sqrt{(b-6)^2}-5|b|=\\\\=|\underbrace {b+5}_{\geq 0}|-|\underbrace {b-6}_{$